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I'm working on the part (b) of the exercise at P.67 of Brown's Cohomology of Groups, which is stated as follows:

Let $R = \mathbb{Z}/n\mathbb{Z}$. Show that $R$ is an injective $R$-module. Deduce:

(a) If $A$ is an abelian group such that $nA = 0$, and $C\subseteq A$ is a cyclic subgroup of order $n$, then $C$ is a direct summand of $A$.

(b) If $A$ is as in (a), then $A$ is a direct sum of cyclic groups.

There are some proofs of (b) which is irrelevant to (a), i.e. in chapter 10 of "Fundamentals of the Theory of Groups" by Mikhail Ivanovich Kargapolov and Ju. I. Merzljakov, though I think these are not the type of proofs that Brown wants us to provide.

After some searching, I find Wofsey's answer to be helpful, but I'm stuck at some point:

(We may assume $n=p^k$ for some prime $p$ if necessary.) Since every cyclic subgroup of order $n$ of such $A$ is a direct summand, by transfinite induction, we may find a subgroup $B\subseteq A$ such that $B$ is the direct sum of copies $\mathbb{Z}/n\mathbb{Z}$ and that every element of $A\backslash B$ has order less than $n$. The author claims that $B$ is a direct summand of $A$, but I can't find out why it holds. I think it suffices to show the following statement, but I can't go any further.

Statement Let $A$ be an abelian group, and let $\{B_m\}$, $\{C_m\}$, $\{C_m'\}$ be subgroups of $A$ with $m\geq 1$, such that

  • $B_1\oplus C_1=A$,
  • $C_m = C_{m+1}\oplus C_{m+1}'$ for $m\geq 1$,
  • $B_{m+1}=B_m\oplus C_{m+1}'$ for $m\geq 1$.

Then $\bigcup_m B_m$ is a direct summand of $A$.

Any help is appreciated.

P.S. This IS my homework assignment, but I've already submitted my answer. So grade is not relevant.

Edit 1: Previous statement is wrong. If we take

  • $A=\prod_{n=1}^{\infty}\mathbb{Z}$,
  • $B_m=\mathbb{Z}\times\dotsb\times\mathbb{Z}\times 0\times 0\times\dotsb$ ($m$ copies of $\mathbb{Z}$),
  • $C_m=0\times\dotsb\times 0\times\mathbb{Z}\times\mathbb{Z}\times\dotsb$ ($m$ copies of $0$),
  • $C_m'=0\times\dotsb\times 0\times\mathbb{Z}\times 0\times 0\times\dotsb$ ($\mathbb{Z}$ appears as the $m$-th term),

then $\bigcup B_m=\bigoplus_{n=1}^\infty\mathbb{Z}$ is not a direct summand of $A$ according to this post.

Noto_Ootori
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1 Answers1

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This was in fact an oversight in my answer (which I have now corrected). You need to use the fact that not only is $\mathbb{Z}/n\mathbb{Z}$ injective as a module over itself, but any direct sum of copies of $\mathbb{Z}/n\mathbb{Z}$ is also injective. (This can be proved directly using Baer's criterion in the same way as for $\mathbb{Z}/n\mathbb{Z}$ itself, or alternatively you can use the fact that a direct sum of injective modules over a Noetherian ring is injective.) That way, at limit steps of the transfinite process you can be sure you still have a direct summand.

(By the way, the condition to stop the process is not that $A\setminus B$ has no elements of order $n$ but that the quotient $A/B$ (which will be isomorphic to any direct complement of $B$ in $A$) has no elements of order $n$.)

Eric Wofsey
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